If someone sees this, can you kinda run thru it in a reply. Its solve using elimination btw.
1 answer:
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Problem 1
w = width
w+36 = length, because it's 36 feet longer compared to the width
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Problem 2
The area w(w+36) is less than 2040, and it's also larger than 0.
We can write that as which is the same as
If you wanted to drop the first part, then you can say either or
I used the formula area = length*width
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Problem 3
There are many possibilities here. Let's say w = 10 feet. If so then the length would be w+36 = 10+36 = 46 feet. This 10 by 46 rectangle has an area of 10*46 = 460 square feet which is under the 2040 sq ft limit.
Another possibility is that w = 20 ft and w+36 = 56. This 20 by 56 rectangle has an area of 20*56 = 1120 sq ft.
It turns out you can pick any value of w between 1 and 30, assuming you only restrict yourself to integers. You can find the largest possible value of w by solving the equation w(w+36) = 2040. The positive solution to this equation is roughly w = 30.62
Side note: even though something like w = 1 is possible, it's not very realistic. A conference hall that's only 1 ft wide won't be able to fit a person comfortably unless they don't mind being sandwiched between two walls and have to walk sideways.
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Problem 4
If you solved w(w+36) = 360 with a graphing calculator or the quadratic formula, then you would find the positive solution for w is roughly 8.15
If your teacher is considering positive real numbers, then this is realistic; however, if they are only considering positive integers, then something like w = 8.15 isn't possible.